Question

In vinyl acetylene $CH\equiv C-CH=C{H}_2$ type of overlapping in $\left(C_2\sigma C_3\right)$ bond is
A.$s{p}^2-sp$
B.$sp-s{p}^2$
C.$s{p}^3-s{p}^3$
D.$s{p}^3-s{p}^2$

Answer 1

Hint:We have to take into consideration the number of bonds formed by each carbon and determine whether the bonds are $\sigma -$bonds (or) $\pi -$bonds. We can determine the hybridization of each atom of carbon using the $\sigma -$bonds (or) $\pi -$bonds.

Complete step by step answer:
We have to first identify the name of atoms of carbon. We can give the structure as,
$$\underset{1}{C}H\equiv \underset{2}{C}-\underset{3}{C}H=\underset{4}{C}{H}_2$$
We have to find the hybridization of two carbons present in the middle. Let us carbon-2.
From the written structure, we could see that it is forming one $\sigma -$bond and two $\pi -$bonds with carbon-3. We can also see it does not form any bonds with hydrogen atoms.
Therefore, it contains two $\sigma -$bonds. We have to consider the bond pairs while determining the hybridization of the atoms. We should not consider the $\pi -$bonds while determining the hybridization of the atom.
We can say that there are two bond pairs involved, and therefore, 2 hybridized orbitals are needed to accommodate them. Hence, the hybridization of this carbon atom will be $sp$.
Let us now consider carbon-3.
From the structure, we could see that one $\sigma -$bond and two $\pi -$bonds with carbon-2. We can also see it forms one $\sigma -$bond with a hydrogen atom.
Let us consider the number of $\sigma -$bond to calculate the hybridization of the carbon. There are 3 $\sigma -$bonds present and 3 hybridized orbitals are needed to accommodate 3 bond pairs. Therefore, the hybridization of the carbon atom will be $s{p}^2$.
Therefore, the option (B) is correct.

Note: We have to be careful while calculating the number of $\sigma -$bond which is formed by each of the carbon and the hydrogen atom since they are not explicitly shown and there is a chance one might miss them. We can prepare vinyl acetylene by Hofmann elimination of the quaternary ammonium salt. We can also prepare using dehydrohalogenation of 1,3-dichloro-2-butene.